Proving the continuity of a function with two variables using partial derivatives

Question

Let

$$f(x,y) = \left\{\begin{array}{rcl}\frac{x^{\alpha}y^{\beta}}{x^2+y^2} & \mathrm{if} & (x,y) \ne (0,0)\\ 0 & \mathrm{if} & (x,y)=(0,0)\end{array}\right.$$

The question is: for what values of $\alpha$ and $\beta$ is $f$ continuous at $(0,0)?$

My (partial) answer is below.

Answer 1

Using polar coordinates ( $x=r\cos \theta,\, y = r\sin \theta $ ) we have

$$\lim_{r\to 0} \frac{r^{\alpha+\beta} \cos^{\alpha} \theta \sin^{\beta} \theta }{r^2} = \lim_{r\to 0 }\, r^{\alpha+\beta-2} \cos^{\alpha} \theta \sin^{\beta} \theta = 0 $$

if

$$ \alpha+\beta - 2> 0 \implies \alpha + \beta > 2. $$

Note:

$$ |\cos^{\alpha} \theta \sin^{\beta} \theta| \leq 1 . $$

Answer 2

I found the answer.

We have $|x^\alpha|\le \(sqrt{(x^2+y^2)})^alpha$ and similarly for $|y^\beta|$.

So $$|x^\alpha y^\beta|\le (x^2+y^2)^{(\alpha+\beta)/2}$$ $$|f(x,y)|\le (x^2+y^2)^{(\alpha+\beta)/2-1}$$ so $\lim_{(x,y)\to (0,0)} f(x,y)$ is smaller than $\lim_{(x,y)\to (0,0)} (x^2+y^2)^{(\alpha+\beta)/2-1}$.

So we make now discussion of the limit according to the values of $\alpha$ and $\beta$ i.e. if $\alpha +\beta <2$ so $(x^2+y^2)^{(\alpha+\beta)/2-1}$ tends to infinity so take $y=x$ then $\lim f(x,x) =1/2$ as $x$ tends to $0$. So it is different from $0$... so f is not cont. At 0 in this case

And so on for $\alpha +\beta >2$ and for $\alpha +\beta =2$ similarly we solve it.